Question

Two vectors
a⃗
and
b⃗
have components, in arbitrary units,
ax=3.2
,
ay=1.6
,
bx=0.5
,
by=4.5
. Find the angle between
a⃗
and
b⃗


33o

28o

57o

62o

Answer 1

Answer on Question #52039 — Physics - Quantum Mechanics

Find angle between two vectors with coordinates:

$$a _ {x} = 3.2, a _ {y} = 1.6, b _ {x} = 0.5, b _ {y} = 4.5$$

Solution:

We have two vectors $\vec{a},\vec{b}$ with components:

$$\vec{a} = (3.2, 1.6,)$$ $$\vec{b} = (0.5, 4.5)$$

Acording to definition of dot product :

$$\vec{a} \cdot \vec{b} = a _ {x} b _ {x} + a _ {y} b _ {y} = | \vec{a} | | \vec{b} | \cos (\phi) \text{, where}$$ $$a _ {x}, a _ {y} \text{ - components of vector } \vec{a}, \quad b _ {x}, b _ {y} \text{ - components of vector } \vec{b}$$ $$| \vec{a} | \text{ module of vector overline } \vec{a}, \quad | \vec{b} | \text{ module of vector } \vec{b}.$$ $$\text{is angle between } \vec{a} \text{ and } \vec{b}.$$

We can find angle between two vectors by next formula:

$$\cos (\phi) = \frac {\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |},$$ $$\vec{a} \cdot \vec{b} = a _ {x} b _ {x} + a _ {y} b _ {y} = 3.2 \cdot 0.5 + 1.6 \cdot 4.5 = 1.6 + 7.2 = 8.8$$ $$| \vec{a} | = \sqrt {a _ {x} ^ {2} + a _ {y} ^ {2}} = \sqrt {3.2 ^ {2} + 1.6 ^ {2}} = \sqrt {10.24 + 2.56} = \sqrt {12.8} = 3.57$$ $$| \vec{b} | = \sqrt {b _ {x} ^ {2} + b _ {y} ^ {2}} = \sqrt {0.5 ^ {2} + 4.5 ^ {2}} = \sqrt {0.25 + 20.25} = \sqrt {20.5} = 4.52$$ $$\cos (\phi) = \frac {\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |} = \frac {8.8}{3.57 \cdot 4.52} = 0.545$$ $$\varphi = \arccos (0.545) = 56.98{}^{\circ}$$

Answer: $\varphi = 57{}^{\circ}$

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